Consider the following. xy dA, D is enclosed by the curves y = x², y = 4x Express D as a region of type 1. {(x, y) | 4 ≤ x ≤ 16, x² sys D = 16x} D = {(x, y) |0 ≤ x ≤ 4, x² ≤ y ≤ 4x} {(x, y) | 0 ≤ x ≤ 16, x² ≤ y ≤ D= 16x} D = ((x, y) |0 ≤ x ≤y, x² ≤ y ≤ 4} D = {(x, y) |0 ≤ x ≤ 4, x² ≤ y ≤ 4x} Express D as a region of type II. D = {(x, y) 10 ≤ y ≤ 4,0 ≤ x ≤ 16} = {(x, y) 1 0 ≤ y ≤ 16, ¼ ≤ x ≤ √y} 4 D = {(x, y) 1 0 ≤ y ≤ 4, 1/1 ≤ √x = √y} ≤√y} D= D = {(x, y) 1 0 ≤ y ≤ 16, 1 ≤ ≤x≤ √Y} D >= {(x, y) 10 ≤ y ≤ 4, 1 y 16 16 Evaluate the double integral in two ways. 512 3

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter9: Multivariable Calculus
Section9.3: Maxima And Minima
Problem 33E
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Consider the following.
Jxy
xy dA,
D is enclosed by the curves y = x², y = 4x
Express D as a region of type I.
D = {(x, y) | 4 ≤ x ≤ 16, x² ≤ y ≤
16x}
D = {(x, y) | 0 ≤x≤ 4, x² ≤ y ≤ 4x}
D = {(x, y) | 0≤x≤ 16, x² ≤ y ≤
16x}
D = {(x, y) | 0 ≤x≤y, x² < y ≤ 4}
D = {(x, y) | 0 ≤ x ≤ 4, x² < y ≤ 4x}
D=
Express D as a region of type II.
D = {(x, y) 1 0 ≤ y ≤ 4,0 ≤ x ≤ 16}
OD = {(x, y) 1 0 ≤ y ≤ 16, ¼ ≤x≤ √y}
Do-{(x10sysa s√x = √y}
X
D=
D = {(x, y) 1 0 ≤ y ≤ 16, 1 ≤ ( = √Y}
S
S
4 16
y
ⒸD= {(x, m) 105 vs 4, 165x5 √y}
Evaluate the double integral in two ways.
512
3
Transcribed Image Text:Consider the following. Jxy xy dA, D is enclosed by the curves y = x², y = 4x Express D as a region of type I. D = {(x, y) | 4 ≤ x ≤ 16, x² ≤ y ≤ 16x} D = {(x, y) | 0 ≤x≤ 4, x² ≤ y ≤ 4x} D = {(x, y) | 0≤x≤ 16, x² ≤ y ≤ 16x} D = {(x, y) | 0 ≤x≤y, x² < y ≤ 4} D = {(x, y) | 0 ≤ x ≤ 4, x² < y ≤ 4x} D= Express D as a region of type II. D = {(x, y) 1 0 ≤ y ≤ 4,0 ≤ x ≤ 16} OD = {(x, y) 1 0 ≤ y ≤ 16, ¼ ≤x≤ √y} Do-{(x10sysa s√x = √y} X D= D = {(x, y) 1 0 ≤ y ≤ 16, 1 ≤ ( = √Y} S S 4 16 y ⒸD= {(x, m) 105 vs 4, 165x5 √y} Evaluate the double integral in two ways. 512 3
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